Question

A football is kicked into the air from an initial height of 4 feet. The height, in feet, of the football above the ground is given by s(t)=-16t^2+50t+4, where t is time, in sevonds, and t >\= 0. At what time will the football be 25 feet above the ground?

A. 3.625 seconds
B. 3.20 seconds
C. .5 seconds or 3.625 seconds
D. .5 seconds or 2.625 seconds

Answer 1

25 = -16t^2 + 50t + 4
16t^2 - 50t + 21 = 0
16t^2 - 8t - 42t + 21 = 0
8t(2t - 1) - 21(2t - 1) = 0
(8t - 21)(2t - 1) = 0
8t - 21 = 0 or 2t - 1 = 0
t = 21/8 or t = 1/2
t = 2.625 or t = 0.5

Answer 2

s ( t ) = - 16 t² + 50 t + 4
25 = - 16 t² + 50 t + 4
16 t² - 50 t + 21 = 0
$t_{12}= \frac{50+/- \sqrt{2500-1344} }{32}= \\ =\frac{50+/-34}{32}$
t 1 = 2.625 s
t 2 = .5 s
Answer: D ) .5 seconds or 2.625 seconds